A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 4 years ago

A 661-N force is being applied to a 121-kg block. The coefficient of static friction between the block and the surface is 0.410. What is the minimal amount of additional mass that must be added on top of the block to prevent the block from moving?

  • This Question is Closed
  1. BTaylor
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The frictional force is the coefficient of friction times the normal force. Here, you want the frictional force to equal 661, so there is no movement. So:\[F_{fr} = \mu_s \times F_N \rightarrow 661 = 0.410 \times mg\] From this, you can solve for mg, and you get mg = 1612 dividing by g (9.8), you get m = 164.5. So, since you already have 121 kg, you only need 43.5 more.

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...


  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.